Problem 103 Binary Tree Zigzag Level Order Traversal
Table of Contents
Problem Statement
Link - Problem 103
Question
Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).
Example 1
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]
Example 2
Input: root = [1]
Output: [[1]]
Example 3
Input: root = []
Output: []
Constraints
- The number of nodes in the tree is in the range `[0, 2000]`.
- `-100 <= Node.val <= 100`
Solution
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
if(root==nullptr)
return {};
vector<vector<int>>ans;
queue<TreeNode*>q;
q.push(root);
bool flag = true;
while(!q.empty()){
int size = q.size();
vector<int>v;
for(int i =0 ;i<size;i++){
TreeNode* t = q.front();
q.pop();
if(t->left)
q.push(t->left);
if(t->right)
q.push(t->right);
if(flag){
v.push_back(t->val);
}
else{
v.insert(v.begin(),t->val);
}
}
ans.push_back(v);
flag =!flag;
}
return ans;
}
};
Complexity Analysis
| Algorithm | Time Complexity | Space Complexity |
| --------------- | --------------- | ---------------- |
| BFS level order | O(n) | O(n) |
Explanation
1. Intuition
- The logic is follows.
- We will do just BFS, but maintain a flag to check if we need to reverse the order of the elements in the vector.
2. Implementation
- If the root is `nullptr` return empty vector.
- Create a vector of vector of int `ans`.
- Create a queue of TreeNode\* `q`.
- Push the root into the queue.
- Create a flag `flag` to check if we need to reverse the order of the elements in the vector.
- Until the queue is not empty
- Get the size of the queue.
- Create a vector of int `v`.
- Loop through the size of the queue
- Get the front element of the queue.
- Pop the front element of the queue.
- If the left of the front element is not `nullptr` push it into the queue.
- If the right of the front element is not `nullptr` push it into the queue.
- If the flag is true push the value of the front element into the vector.
- Else insert the value of the front element at the beginning of the vector.
- Push the vector into the `ans`.
- Reverse the flag.
- Return the `ans`.